From 2137778d44d256795bd0ca6c9b880b2c3be180d1 Mon Sep 17 00:00:00 2001
From: damon <245211612@qq.com>
Date: Thu, 26 May 2022 00:29:02 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880015.=E4=B8=89?=
 =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C.md=EF=BC=89=EF=BC=9A=E5=A2=9E?=
 =?UTF-8?q?=E5=8A=A0javascript=E7=89=88=E6=9C=ACnsum=E7=9A=84=E9=80=9A?=
 =?UTF-8?q?=E7=94=A8=E8=A7=A3=E6=B3=95?=
MIME-Version: 1.0
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---
 problems/0015.三数之和.md | 70 +++++++++++++++++++++++++++++++++++
 1 file changed, 70 insertions(+)

diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index cc184c87..12d83d8f 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -345,6 +345,76 @@ var threeSum = function(nums) {
     return res
 };
 ```
+
+解法二：nSum通用解法。递归
+
+```js
+/**
+ *  nsum通用解法，支持2sum，3sum，4sum...等等
+ *  时间复杂度分析：
+ *  1. n = 2时，时间复杂度O(NlogN)，排序所消耗的时间。、
+ *  2. n > 2时，时间复杂度为O(N^n-1)，即N的n-1次方，至少是2次方，此时可省略排序所消耗的时间。举例：3sum为O(n^2)，4sum为O(n^3)
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var threeSum = function (nums) {
+    // nsum通用解法核心方法
+    function nSumTarget(nums, n, start, target) {
+        // 前提：nums要先排序好
+        let res = [];
+        if (n === 2) {
+            res = towSumTarget(nums, start, target);
+        } else {
+            for (let i = start; i < nums.length; i++) {
+                // 递归求(n - 1)sum
+                let subRes = nSumTarget(
+                    nums,
+                    n - 1,
+                    i + 1,
+                    target - nums[i]
+                );
+                for (let j = 0; j < subRes.length; j++) {
+                    res.push([nums[i], ...subRes[j]]);
+                }
+                // 跳过相同元素
+                while (nums[i] === nums[i + 1]) i++;
+            }
+        }
+        return res;
+    }
+
+    function towSumTarget(nums, start, target) {
+        // 前提：nums要先排序好
+        let res = [];
+        let len = nums.length;
+        let left = start;
+        let right = len - 1;
+        while (left < right) {
+            let sum = nums[left] + nums[right];
+            if (sum < target) {
+                while (nums[left] === nums[left + 1]) left++;
+                left++;
+            } else if (sum > target) {
+                while (nums[right] === nums[right - 1]) right--;
+                right--;
+            } else {
+                // 相等
+                res.push([nums[left], nums[right]]);
+                // 跳过相同元素
+                while (nums[left] === nums[left + 1]) left++;
+                while (nums[right] === nums[right - 1]) right--;
+                left++;
+                right--;
+            }
+        }
+        return res;
+    }
+    nums.sort((a, b) => a - b);
+    // n = 3，此时求3sum之和
+    return nSumTarget(nums, 3, 0, 0);
+};
+```
+
 TypeScript:
 
 ```typescript

From 4e95d663a047d1745c57b3f46cba055836e8ec13 Mon Sep 17 00:00:00 2001
From: damon <245211612@qq.com>
Date: Fri, 27 May 2022 22:52:57 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880509.=E6=96=90?=
 =?UTF-8?q?=E6=B3=A2=E9=82=A3=E5=A5=91=E6=95=B0.md=EF=BC=89=EF=BC=9A?=
 =?UTF-8?q?=E5=A2=9E=E5=8A=A0javascript=E7=89=88=E6=9C=AC=E4=BC=98?=
 =?UTF-8?q?=E5=8C=96=E7=A9=BA=E9=97=B4=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=B0?=
 =?UTF-8?q?O(1)=E8=A7=A3=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0509.斐波那契数.md | 18 ++++++++++++++++++
 1 file changed, 18 insertions(+)

diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md
index 1d17784d..7c899195 100644
--- a/problems/0509.斐波那契数.md
+++ b/problems/0509.斐波那契数.md
@@ -234,6 +234,7 @@ func fib(n int) int {
 }
 ```
 ### Javascript 
+解法一
 ```Javascript
 var fib = function(n) {
     let dp = [0, 1]
@@ -244,6 +245,23 @@ var fib = function(n) {
     return dp[n]
 };
 ```
+解法二：时间复杂度O(N)，空间复杂度O(1)
+```Javascript
+var fib = function(n) {
+    // 动规状态转移中，当前结果只依赖前两个元素的结果，所以只要两个变量代替dp数组记录状态过程。将空间复杂度降到O(1)
+    let pre1 = 1
+    let pre2 = 0
+    let temp
+    if (n === 0) return 0
+    if (n === 1) return 1
+    for(let i = 2; i <= n; i++) {
+        temp = pre1
+        pre1 = pre1 + pre2
+        pre2 = temp
+    }
+    return pre1
+};
+```
 
 TypeScript